左旋转移位(程序填空)求助

Question

利用C++语言的移位运算实现左旋转移位。
#include
//val左旋转移n位，1<=n<=7
unsigned char rol(unsigned char val,int n);
//val左旋转移位(rol函数被重载)
unsigned char rol(unsigned char val);
//输出val对应的二进制值
void showBit(unsigned char val);
int main() {
unsigned char x=0xb5;
showBit(x); //输出x对应的二进制值

showBit(rol(x)); //输出x左移1位后对应的二进制值

int n;
std::cin>>n;
showBit(rol(x,n)); //输出x左移n位后对应的二进制值

return 0;

}
unsigned char rol(unsigned char x,int n) {

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}
unsigned char rol(unsigned char val) {

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}
void showBit(unsigned char val) {

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}
输入
1<=正整数<=7
输出
10110101B 01101011B 左移n位后对应的二进制值B

Answer 1

问题解决请点采纳，包括前面几个问题

unsigned char rol(unsigned char val) {
    return (val % 128) * 2 + val / 128;
}

unsigned char rol(unsigned char x,int n) {
    for (int i = 0; i < n; i++, x = rol(x));
    return x;
}

void showBit(unsigned char val, int x = 8) {
    if (x == 0) 
    {
        cout << endl;
        return;
    }
    showBit(val / 2, x - 1);
    cout << val % 2;
}